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3 years ago

Which unit of measurement is included in the International System of Unit! (si)?

Physics

1 answer:

ozzi3 years ago

3 0

Answer:

See the explanation below

Explanation:

There are several measures for the international system of measures. Let's name some and their representation symbol.

meter = [m]

time = [s] = seconds

mass = [kg] = kilograms

Temperature = [°C] = celcius degrees

Power = [W] = watts.

Force = [N] = Newtons

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Marking brainliest help pls the formula are there to help ^

natta225 [31]

Answer:

Explanation:

Look at the equation for Potential Energy. PE = mass times gravity times the height. Filling in and solving for h:

34.3 = .5(9.8)h so

34.3 = 4.9h so

h = 7 meters

6 0

3 years ago

An unbanked circular highway curve on level ground makes aturn

tino4ka555 [31]

Answer:

Explanation:

Given

Velocity of traffic $v=60\ mi/hr\approx 26.82\ m/s$

Maximum value of Centripetal force is one-tenth of weight

such that $(F_c)_{max}=\frac{mg}{10}$

to make a safe turn centripetal force with radius of curvature r is given by

$F_c=\frac{mv^2}{r}$

$(F_c)_{max}=\frac{mg}{10}=F_c=\frac{mv^2}{r}$

$\frac{mg}{10}=\frac{mv^2}{r}$

$r=\frac{10v^2}{g}$

$r=\frac{10\times 26.82^2}{9.8}$

$r=733.99\approx 734\ m$

4 0

4 years ago

Two particles with charges of 5.00 μ C and -3.00 μC are placed 0.250 m apart. Where can a third charge be placed so that the net

MAXImum [283]

Answer:

0.86 m

Explanation:

q₁ = magnitude of positive charge = 5 x 10⁻⁶ C

q₂ = magnitude of negative charge = 3 x 10⁻⁶ C

r = distance between the two charges = 0.250 m

d = distance of the location of third charge from negative charge

q = magnitude of charge on third charge

Using equilibrium of electric force on third charge

$\frac{kq_{2}q}{d^{2}} = \frac{kq_{1}q}{(r+d)^{2}}$

$\frac{q_{2}}{d^{2}} = \frac{q_{1}}{(r+d)^{2}}$

$\frac{(5\times 10^{-6})}{(0.250+d)^{2}} = \frac{(3\times 10^{-6})}{(d^{2}}$

d = 0.86 m

4 0

3 years ago

Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g

tekilochka [14]

Answer: a) It will take more time to return to the point from which it was released

Explanation: To determine how long it takes for the ball to return to the point of release and considering it is a free fall system, we can use the given formula:

$d=v_{0}.t + \frac{1}{2} .a.t^{2}$ , where:

d is the distance the ball go through;

v₀ is the initial velocity, which is this case is 0 because he releases the ball;

a is acceleration due to gravity;

t is the time necessary for the fall;

Suppose <em>h</em> is the height from where the ball was dropped.

On Earth:

h=0.t + $\frac{1}{2}.10.t^{2}$

h = 5t²

$t_{T}$ = $\sqrt{\frac{h}{5} }$

On the other planet:

h = 0.t + $\frac{1}{2}.30.t^{2}$

h = 15.t²

$t_{P}$ = $\sqrt{\frac{h}{15} }$

Comparing the 2 planets:

$\frac{t_{T} }{t_{P} } = \frac{\sqrt{\frac{h}{5} } }\sqrt{{\frac{h}{15} } }$

$\frac{t_{T} }{t_{P} }$ = $\sqrt{3}$ or $t_{T} = \sqrt{3}.t_{P}$

Comparing the two planets, on the massive planet, it will take more time to fall the height than on Earth. In consequence, it will take more time to return to the initial point, when it was released.

5 0

3 years ago

A microscope has an objective lens with a focal length of 14.0mm . A small object is placed 0.80mm beyond the focal point of the

Nikolay [14]

Answer:

A. 260 mm

B. - 18

C. 175

Explanation:

The expression for the lens equation is

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$